Question

After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent...

After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on the Tukey procedure. At α = .05, if the confidence interval includes the value of zero, then we can reject the hypothesis that the two population means are equal.

True

False

The error sum of squares measures the between-treatment variability.

True

False

The experimentwise α for the 95 percent individual confidence interval for μ1μ2 (treatment mean 1 − treatment mean 2) will always be smaller than the experimentwise α for a Tukey 95 percent simultaneous confidence interval for μ1μ2.

True

False

In one-way ANOVA, a large value of F results when the within-treatment variability is large compared to the between-treatment variability.

True

False

Homework Answers

Answer #1

False , this is not true that After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on the Tukey procedure. At α = .05, if the confidence interval includes the value of zero, then we can reject the hypothesis that the two population means are equal.

True, The error sum of squares measures the between-treatment variability.

this is True , The experimentwise α for the 95 percent individual confidence interval for μ1μ2 (treatment mean 1 − treatment mean 2) will always be smaller than the experimentwise α for a Tukey 95 percent simultaneous confidence interval for μ1μ2.

true ,In one-way ANOVA, a large value of F results when the within-treatment variability is large compared to the between-treatment variability.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32...
The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 46 33 30 45 36 30 46 35 26 48 36 32 50 40 Sample mean 30 47 36 Sample variance 6 4 6.5 At the  = .05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares,...
In an experiment designed to test the output levels of three different treatments, the following results...
In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST = 320, SSTR = 130, nT = 19. Set up the ANOVA table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Error Total Test for any significant difference between the mean output levels of the three treatments....
The following data were obtained for a randomized block design involving five treatments and three blocks:...
The following data were obtained for a randomized block design involving five treatments and three blocks: SST = 510, SSTR = 370, SSBL = 95. Set up the ANOVA table. (Round your value for F to two decimal places, and your p-value to three decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Blocks Error Total Test for any significant differences. Use α = 0.05. State the null and alternative hypotheses. H0: Not...
Question 14 For all tests of hypothesis, the probability of rejecting the null hypothesis is a...
Question 14 For all tests of hypothesis, the probability of rejecting the null hypothesis is a function of A. the size of the observed differences B. the alpha level and the use of one- or two-tailed tests C. sample size D. all of the above Question 15 The Central Limit Theorem states that as sample size becomes large the population distribution becomes normal. True False Question 16 The area between the mean and a Z score of + 1.50 is...
Treatment A Treatment B Treatment C 32 44 34 30 43 37 30 44 36 26...
Treatment A Treatment B Treatment C 32 44 34 30 43 37 30 44 36 26 46 37 32 48 41 Sample mean 30 45 37 Sample variance 6 4 6.5 a. At the X= 0.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error Calculate the value...
ANOVA calculations and rejection of the null hypothesis The following table summarizes the results of a...
ANOVA calculations and rejection of the null hypothesis The following table summarizes the results of a study on SAT prep courses, comparing SAT scores of students in a private preparation class, a high school preparation class, and no preparation class. Use the information from the table to answer the remaining questions. Treatment Number of Observations Sample Mean Sum of Squares (SS) Private prep class 60 650 132,750.00 High school prep class 60 645 147,500.00 No prep class 60 625 162,250.00...
Given the following sample information, test the hypothesis that the treatment means are equal at the...
Given the following sample information, test the hypothesis that the treatment means are equal at the 0.05 significance level. Treatment 1 Treatment 2 Treatment 3 3 9 6 2 6 3 5 5 5 1 6 5 3 8 5 1 5 4 4 1 7 4 6 4 (a) State the null hypothesis and the alternate hypothesis. Ho : ?1 (>, <, or =) incorrect ?2 = (<,>, or =) ?3. H1 : Treatment means are not ___ all...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II III n = 6 n = 4 n = 4                    M = 2 M = 2.5 M = 5 N = 14 T = 12 T = 10 T = 20 G = 42 SS = 14 SS = 9 SS = 10 ΣX2tot = 182 Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the...
Null hypothesis (H0): The true difference in mean hemoglobin change for pregnant women not exposed to...
Null hypothesis (H0): The true difference in mean hemoglobin change for pregnant women not exposed to herbicides versus those who are somewhat exposed is 0 g/dL. (µbottle – µboth = 0 g/dL) Research hypothesis (H1): The true difference in mean hemoglobin change for pregnant women not exposed to herbicides versus those who are somewhat exposed is not 0 g/dL. (µbottle – µboth ≠ 0 g/dL) SAS Output: The TTEST Procedure Variable: change group Method Mean 95% CL Mean Std Dev...
T/F Question and explain 1.A 95% confidence interval for population mean μ is 65.6±12.8 from a...
T/F Question and explain 1.A 95% confidence interval for population mean μ is 65.6±12.8 from a sample of size n=96. If one took a second random sample of the same size, then the probability that the 95% confidence interval for μ based on the second sample contains 65.6 is 0.95. 2.The probability of a Type I error when α=0.05 and the null hypothesis is true is 0.05. 3.Because an assumption of ANOVA is that all of the population variances are...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT